According to this post, we found that a $n$ times integrated Brownian motion could be expressed as, \begin{align} V_n(t) = \int_0^t V_{n-1}(s)\ ds = \frac{1}{n!} \int_0^t (t-s)^n\ dB_s, \end{align} where $V_1(t) = \int_0^t B_s\ ds$. Using the above result, I want to prove that the process \begin{align} M_n(t) = V_n(t) - \sum_{j=1}^n \frac{t^j}{j!\cdot (n-j)!} \int_0^t (-s)^{n-j}\ dB_s \end{align} is a martingale.
Therefore, I interpret $M_n(t) = f(t,B_t)$ as a function of $t, B_t$. Now, if \begin{align} \int_0^T \mathbb{E}\big[f_{B_t}(t,B_t)^2\big] ds < \infty \end{align} and $f_t + \frac{1}{2}f_{B_t,B_t} = 0$ then $f(t,B_t)$ is an martingale on $[0,T]$.
Now, I am wondering how to differentiate \begin{align} &\frac{d}{dt} V_n(t) = \frac{1}{n!} \int_0^t \frac{d}{dt}\big[(t-s)^n\ \big]dB_s \qquad \text{and} \\ &\frac{d}{dt} \bigg[ \sum_{j=1}^n \frac{t^j}{j!\cdot (n-j)!} \int_0^t (-s)^{n-j}\ dB_s \bigg], \end{align} since $t$ is in the integrand of the integral as well. Or should I try a whole different approach?
Edit: \begin{align} M_n(t) &= V_n(t) - \sum_{j=1}^n \frac{t^j}{j!\cdot (n-j)!} \int_0^t (-s)^{n-j}\ dB_s \\ &= \frac{1}{n!} \int_0^t (t-s)^n\ dB_s - \sum_{j=1}^n \frac{t^j}{j!\cdot (n-j)!} \int_0^t (-s)^{n-j}\ dB_s \\ &= \frac{1}{n!} \int_0^t \sum_{j=0}^n {n \choose j} t^j (-s)^{n-j} \ dB_s - \sum_{j=1}^n \frac{t^j}{j!\cdot (n-j)!} \int_0^t (-s)^{n-j}\ dB_s \\ &= \int_0^t \sum_{j=0}^n \frac{t^j}{j!\cdot (n-j)!} (-s)^{n-j}\ dB_s - \int_0^t \sum_{j=1}^n \frac{t^j}{j!\cdot (n-j)!}(-s)^{n-j}\ dB_s \\ &= \frac{1}{n!} \int_0^t (-s)^n\ dB_s. \end{align}